Tag: maths

Questions Related to maths

The locus of a point whose chord of contact to the ellipse $x^{2}+2y^{2}=1$ subtends a right angle at the centre of the ellipese is 

position of point wrt ellipse ellipse maths

  1. $x^{2}+4y^{2}=3$

  2. $y^{2}=4x$

  3. $2x^{2}+y^{2}=1$

  4. none of these

Show answer
Correct Option: A
Explanation:

Equation of ellipse-

${x}^{2} + 2 {y}^{2} = 1 ..... \; \left(  1 \right)$
Let $\left( h, k \right)$ be the point whose chord of contact subtends a right angle at the centre of ellipse.
Equation of chord of contact-
$hx + 2ky = 1$
Squaring both sides, we have
${\left( hx + 2ky \right)}^{2} = {\left( 1 \right)}^{2}$
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = 1 ..... \left( 2 \right)$
Now, from equation $\left( 1 \right) & \left( 2 \right)$, we have
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = {x}^{2} + 2 {y}^{2}$
$\Rightarrow \left( {h}^{2} - 1 \right) {x}^{2} + \left( 4 {k}^{2} - 2 \right) {y}^{2} + 4hkxy = 0$
The above equation represents a pair of perpendicular lines if
Coefficient of ${x}^{2} + $ Coefficient of ${y}^{2} = 0$
$\left( {h}^{2} - 1 \right) + \left( 4 {k}^{2} - 2 \right) = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} - 3 = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} = 3$
Replacing $h$ and $k$ with $x$ and $y$ respectively, we get
${x}^{2} + 4 {y}^{2} = 3$
Hence the locus of the point whose chord of contact subtends a right angle at the centre of ellipse is ${x}^{2} + 4 {y}^{2} = 3$.

Equation of the largest circle with centre (1,0) that can be inscribed in the ellipse $x^2 + 4y^2 = 16$ is 

position of point wrt ellipse ellipse maths

  1. $2x^2 + 2y^2 - 4x + 7 = 0$

  2. $x^2 + y^2 - 2x + 5 = 0$

  3. $3x^2 + 3y^2 - 6x - 8 = 0$

  4. None of these

Show answer
Correct Option: C
Explanation:
$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{4}=1$

Point on the ellipse $(4\cos\theta, 2\sin\theta)$

Let the circle have radius $=r$

$(x-1)^{2}+(y-0)^{2}=r^{2}$ Solving if with ellipse

$x^{2}+4y^{2}=16$

$(x-1)^{2}+\dfrac{(16-x^{2})}{4}=r^{2}$

$4(x^{2}-2x+1)+16-x^{2}=4r^{2}$

$3x^{2}-8x+20-4r^{2}=0$

As the circle & ellipse touch each other 

$D=0$

$8^{2}-4.2\times (20-4r^{2})=0$

$r^{2}=\dfrac{\pi}{3}$

$(x-1)^{2}+y^{2}=\dfrac{11}{3}$

$3x^{2}+3y^{2}-6x-8=0$

An ellipse of major axis $20\sqrt {3}$ and minor axis $20$ slides along the coordinate axes and always remains confined in the $1^{st}$ quadrant. The locus of the centre of the ellipse therefore describes the arc of a circle. The length of this arc is

position of point wrt ellipse ellipse maths

  1. $5\pi$

  2. $20\pi$

  3. $\dfrac {5\pi}{3}$

  4. $\dfrac {20\pi}{3}$

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Correct Option: B

A tangent to the ellipse $4x^2+9y^2=36$ is cut by tangent at the extremities of the major axis at $T$ and $T'$. The circles on $TT'$ as diameters passes through the point 

position of point wrt ellipse ellipse maths

  1. $(0,-\sqrt5)$

  2. $(\sqrt5,0)$

  3. $(0,0)$

  4. $(3,2)$

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Correct Option: B

If the line $x\, cos\, \alpha+y\,sin \,\alpha=p$ is normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then 

position of point wrt ellipse ellipse maths

  1. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=a^2-b^2$

  2. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=(a^2-b^2)^2$

  3. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=a^2-b^2$

  4. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=(a^2-b^2)^2$

Show answer
Correct Option: A

Let $(a, 0)$ and $B(b, 0)$ be fixed distinct points on the $x-axis$, none of which coincides with the origin $O(0, 0)$ and let $C$ be a point on the $y-axis$. Let $L$ be a line through the $O(0, 0)$ and perpendicular to the line $AC$, The locus of the point of intersection of lines $L$ and $BC$ if $C$ varies along the $y-axis$, is (provided $x^{2}+ab\neq 0$) 

position of point wrt ellipse ellipse maths

  1. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=x$

  2. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=y$

  3. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=x$

  4. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=y$

Show answer
Correct Option: D

If P($\theta$) and Q($\pi$/2 + $\theta$) are two points on the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Locus of the mid-point of PQ is

position of point wrt ellipse ellipse maths

  1. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{2}$

  2. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 4$

  3. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2$

  4. None of these

Show answer
Correct Option: A
Explanation:

Given,$P(\theta),Q(\frac{\pi}{2}+\theta)$ are two points on the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$
Any point on the ellipse will be $(acos\theta,bsin\theta)$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(acos(\frac{\pi}{2}+\theta),bsin(\frac{\pi}{2}+\theta))$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(-asin\theta,bcos\theta)$
Let required point be $C(x,y)$
Given, $C=mid-point\;of\;PQ$
$\Rightarrow (x,y)=(\displaystyle\frac{(acos\theta-asin\theta)}{2},\displaystyle\frac{(bsin\theta+bcos\theta)}{2})$
$\Rightarrow \displaystyle\frac{x}{a}=(\displaystyle\frac{(cos\theta-sin\theta)}{2}),\displaystyle\frac{y}{b}=(\displaystyle\frac{(sin\theta+cos\theta)}{2})$
on squaring $\displaystyle\frac{x}{a},\displaystyle\frac{y}{b}$ and adding both
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=(\displaystyle\frac{(cos\theta-sin\theta)}{2})^2+(\displaystyle\frac{(sin\theta+cos\theta)}{2})^2$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{2(cos^2\theta+sin^2\theta)}{4}$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{1}{2}$(since $cos^2\theta+sin^2\theta=1$)

The value of $\alpha$ for which the point $(\alpha,\alpha+2)$ is an interior point of smaller segment of the curve $x^{2}+y^{2}-4=0$ made by the chord of the curve whose equation is $3x+4y+12=0$ is

position of point wrt ellipse ellipse maths

  1. $\left(-\infty,\dfrac {-20}{7}\right)$

  2. $(-2,0)$

  3. $\left(-\infty,\dfrac {20}{7}\right)$

  4. $\alpha\ \epsilon\ \phi$

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Correct Option: A

The distance of a point on the ellipse $\dfrac {x^{2}}{6}+\dfrac {y^{2}}{2}=1$ from the centre is $2$, then the eccentric angle is-

position of point wrt ellipse ellipse maths

  1. $\dfrac \pi3$

  2. $\dfrac \pi4$

  3. $\dfrac \pi6$

  4. $\dfrac \pi2$

Show answer
Correct Option: B
Explanation:
Given ellipse $ \dfrac{x^{2}}{6}+\dfrac{y^{2}}{2} = 1 $

Let $ \theta$ be the eccentric angle of the point $p$

coordinate of $p$ $ (\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$

Given distance $= 2 $

$ \therefore $ $OP = 2$

$ \sqrt{6 cos^{2}\theta +2sin^{2}\theta } = 2 \Rightarrow 6cos^{2}\theta +2sin^{2}\theta  = 4$

$ 3cos^{2}\theta +sin^{2}\theta  = 2 $

$ 2 sin^{2}\theta  = 1$

$ sin^{2}\theta  = \dfrac{1}{2} \Rightarrow  sin\theta  = \pm  \dfrac{1}{\sqrt{2}}$

$ \therefore $ eccentric angle $\theta  = \pm \dfrac{\pi }{4}$

A rod of length $l$ rests against a vertical wall and a floor of a room.Let P be a point on the rod,nearer to its end on the wall, that divides its length in the ratio 1:2 if the rod begins to slide on the floor,then the locus of P is:

position of point wrt ellipse ellipse maths

  1. an ellipse of eccentricity $\dfrac { 1 }{ 2 }$

  2. an ellipse of eccentricity $\dfrac { \sqrt { 3 } }{ 2 }$

  3. a circle of radius $\dfrac { l }{ 2 }$

  4. a circle of radius $\dfrac { \sqrt { 3 } }{ 2 } l$

Show answer
Correct Option: B